Video Transcript
Use Euler’s formula to express sin
cubed 𝜃 cos squared 𝜃 in the form 𝑎 sin 𝜃 plus 𝑏 sin three 𝜃 plus 𝑐 sin five
𝜃, where 𝑎, 𝑏, and 𝑐 are constants to be found. Hence, find the solutions of sin
five 𝜃 minus sin three 𝜃 equals zero in the interval 𝜃 is greater than or equal
to zero and less than 𝜋. Give your answer in exact form.
We begin by recalling the fact that
sin 𝜃 is equal to one over two 𝑖 times 𝑒 to the 𝑖𝜃 minus 𝑒 to the negative
𝑖𝜃. And cos 𝜃 is equal to a half times
𝑒 to the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃. This means we can find the product
of sin cubed 𝜃 and cos squared 𝜃. We can write it as one over two 𝑖
times 𝑒 to the 𝑖𝜃 minus 𝑒 to the negative 𝑖𝜃 cubed times a half times 𝑒 to
the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃 squared. One over two 𝑖 cubed is negative
one over eight 𝑖. And a half squared is
one-quarter. So we can rewrite our expression a
little bit further. We find the product of negative one
over eight 𝑖 and a quarter. And we get negative one over
32𝑖. And we can rewrite the rest of our
expression as shown.
We’re now going to use the binomial
theorem to expand each of the sets of parentheses. The first part becomes 𝑒 to three
𝑖𝜃 plus three choose one 𝑒 to the two 𝑖𝜃 times negative 𝑒 to the negative 𝑖𝜃
and so on. And this simplifies to 𝑒 to the
three 𝑖𝜃 minus three 𝑒 to the 𝑖𝜃 plus three 𝑒 to the negative 𝑖𝜃 minus 𝑒 to
the negative three 𝑖𝜃. Let’s repeat this process for 𝑒 to
the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃 squared. When we do, we get 𝑒 to the two
𝑖𝜃 plus two 𝑒 to the zero which is just two plus 𝑒 to the negative two 𝑖𝜃.
We’re going to need to find the
product of these two expressions. We’ll need to do that really
carefully. We’ll need to ensure that each term
in the first expression is multiplied by each term in the second expression. And we can write sin cubed 𝜃 cos
squared 𝜃 as shown. Now, there’s quite a lot going on
here. So you might wish to pause the
video and double-check your answer against mine. We’re going to gather the
corresponding powers of 𝑒 together.
We’ll gather 𝑒 to the five 𝑖𝜃
and 𝑒 to the negative five 𝑖𝜃. We’ll collect 𝑒 to the plus and
minus three 𝑖𝜃. And we’ll gather 𝑒 the 𝑖𝜃 and 𝑒
to the negative 𝑖𝜃. Let’s neaten things up
somewhat. We end up with negative one over
32𝑖 times 𝑒 to the five 𝑖𝜃 minus 𝑒 to the negative five 𝑖𝜃 minus 𝑒 to the
three 𝑖𝜃 minus 𝑒 to the negative three 𝑖𝜃 minus two times 𝑒 to the 𝑖𝜃 minus
𝑒 to the negative 𝑖𝜃. And now, you might be able to spot
why we chose to do this. We can now go back to the given
formulae. Let’s clear some space for the next
step.
We kind of unfactorize a
little. And we can rewrite sin cubed 𝜃 cos
squared 𝜃 as shown. And we can therefore replace 𝑒 to
the 𝑖𝜃 plus 𝑒 to the negative 𝑖𝜃 with sin 𝜃 and so on. And we can see that sin cubed 𝜃
cos squared 𝜃 is equal to a 16th times two sin 𝜃 plus sin three 𝜃 minus sin five
𝜃. Since 𝑎, 𝑏, and 𝑐 are constants
to be found, we can say that 𝑎, the coefficient of sin 𝜃, is one-eighth. 𝑏, the coefficient of sin three
𝜃, is a 16th. And 𝑐, the coefficient sin five
𝜃, is negative one 16th.
Let’s now consider part two of this
question. We begin by using our answer to
part one and multiplying both sides by 16. We then subtract two sin 𝜃 from
both sides and multiply through by negative one. And we can now see that we’ve got
an equation in sin five 𝜃 minus sin three 𝜃. We’re told that sin five 𝜃 minus
sin three 𝜃 is equal to zero. So we let two sin 𝜃 minus 16 sin
cubed 𝜃 cos squared 𝜃 be equal to zero. And then, we factor by two sin
𝜃. Since the product of these two
terms is equal to zero, this means that either of these terms must be equal to
zero. So either two sin 𝜃 is equal to
zero and dividing by two, we could see that sin 𝜃 is equal to zero or one minus
eight sin squared 𝜃 cos squared 𝜃 is equal to zero. Given the interval 𝜃 is greater
than or equal to zero and less than 𝜋, we can see that one of our solutions is when
𝜃 is equal to zero.
We’re going to rewrite our other
equations somewhat. We know that sin two 𝜃 is equal to
two sin 𝜃 cos 𝜃. Squaring this, we get sin squared
two 𝜃 equals four sin squared 𝜃 cos squared 𝜃. And that’s in turn means that our
equation is one minus two sin squared two 𝜃 equals zero. Rearranging to make sin two 𝜃 the
subject, we see that sin two 𝜃 is equal to plus or minus one over root two. Starting with the positive square
root for 𝜃 in the interval given, we know that sin two 𝜃 is equal to one over root
two when 𝜃 is equal to 𝜋 by eight or three 𝜋 by eight. Similarly, we can solve for the
negative square root. And we get five 𝜋 by eight and
seven 𝜋 by eight. And there are therefore five
solutions to the equation sin five 𝜃 minus sin three 𝜃 equals zero in the interval
𝜃 is greater than or equal to zero and less than 𝜋. They are zero 𝜋 by eight, three 𝜋
by eight, five 𝜋 by eight, and seven 𝜋 by eight.
